Linear Spaces

Span and Linear Independence

Linear Combination: A linear combination of a set of vectors S={v1,v2,…,vn}S = \{v_1, v_2, \dots, v_n\} in a vector space VV is a vector of the form c1v1+c2v2+⋯+cnvnc_1v_1 + c_2v_2 + \dots + c_nv_n, where c1,c2,…,cnc_1, c_2, \dots, c_n are scalars.

Span: Let S={v1,v2,…,vn}S = \{v_1, v_2, \dots, v_n\} be a set of vectors in a vector space VV. The set of all linear combinations of the vectors in SS is called the span of SS and is denoted by Span(S)Span(S).

Linear Independence: A set of vectors S={v1,v2,…,vn}S = \{v_1, v_2, \dots, v_n\} in a vector space VV is called linearly independent iff the only linear combination of the vectors in SS that equals the zero vector is the trivial linear combination, that is, c1v1+c2v2+⋯+cnvn=0c_1v_1 + c_2v_2 + \dots + c_nv_n = 0 implies c1=c2=⋯=cn=0c_1 = c_2 = \dots = c_n = 0.

Zero vector cannot be an element of an independent set.


Example: X={[01],[01]}X = \left\{\begin{bmatrix} 0 \\ 1 \end{bmatrix} ,\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}


Example: Consider the linear space of polynomials with degree n≤2n \leq 2. Let subset S={p1,p2,p3}S=\{p_1, p_2, p_3\} where p1(t)=1p_1(t) = 1, p2(t)=tp_2(t) = t, p3(t)=t2p_3(t) = t^2. Is SS linearly independent?
 ~Proof: Let a1,a2,a3∈Ra_1, a_2, a_3 \in \mathbb{R},


Example: S={cos(t),sin(t),cos(t−π/3)}S =\{cos(t), sin(t), cos(t-\pi/3)\}
 ~Proof: Let a1,a2,a3∈Ra_1, a_2, a_3 \in \mathbb{R},


#EE501 - Linear Systems Theory at METU